Question

If \(N_2\) gas is bubbled through water at 293 K, how many moles of \(N_2\) gas would dissolve in 1 litre of water? Assume that \(N_2\) exerts a partial pressure of 0.987 bar. [Given \(K_H\) for \(N_2\) at 293 K is 76.48 K bar]

• A. \(7.16 \times 10^{-5}\)
• B. \(7.16 \times 10^{-4}\)
• C. \(7.16 \times 10^{-3}\)
• D. \(0.716 \times 10^{-3}\)

Hint:

To calculate the solubility of a gas in water, remember that Henry’s Law is useful when the gas behaves ideally and is at low concentrations.

Answer 1

The Correct Option is B

Using Henry's Law, \[ X = \frac{P}{K_H} = \frac{0.987}{76.48 \times 10^3} \] \[ X = 0.0129 \times 10^{-3} \] Since \(n_{N_2} \ll n_{\text{water}}\), we can use the approximation: \[ n_{N_2} = n_{\text{water}} \times 0.0129 \times 10^{-3} \] Where \(n_{\text{water}} = 1000 \, \text{g}\) (1 litre of water) and the molar mass of water is 18 g/mol. Thus, \[ n_{N_2} = \frac{1000}{18} \times 0.0129 \times 10^{-3} = 0.000716 \, \text{mol} \] Thus, the number of moles of \(N_2\) gas dissolved is \(0.716 \times 10^{-3}\).